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{\bf HW6 Solution}
\begin{itemize}
  \item (HW6\underline{ }1)
        \begin{itemize}
        \item[(a)]Each trail has only two outcomes: success or failure. $p=P(success)$ is the same
                  for every trail. Trails are independent.
        \item[(b)]If we take a sample of size $n$ from a much larger population (of size $N$) in which
                  a proportion $p$ have a characteristic of interest, then the distribution of $X$,
                  the number in the sample with that characteristic, is approximately $Binominal(n,p)$.
                  (from slide 83, class notes, week 6). 
        \item[(c)]$X=$ number of successes in 10 trails. $P(success)=0.3$.
                  \begin{eqnarray*}
                  P(X=x)={10 \choose x}0.3^x(1-0.3)^{10-x}, x=0,1,\cdots,10.
                  \end{eqnarray*} 
        \item[(d)]\begin{eqnarray*}
                  P(A)&=& P(1\leq X <4)\\
                      &=& P(X=1)+P(X=2)+P(X=3)\\
                      &=& {10 \choose 1}0.3^1 0.7^{10-1}+{10 \choose 2}0.3^2 0.7^{10-2}
                          + {10 \choose 3}0.3^3 0.7^{10-3}\\
                      &=& 0.62.
                  \end{eqnarray*}
                  \begin{eqnarray*}
                  P(B)&=& P(4 < X \leq 10)\\
                      &=& P(X=5)+P(X=6)+P(X=7)+P(X=8)+P(X=9)+P(X=10)\\
                      &=& 1-P(X=0)-P(X=1)-P(X=2)-P(X=3)-P(X=4)\\
                      &=& 0.15.
                  \end{eqnarray*} 
                  So, $P(A)$ is bigger.
        \end{itemize}
        
  \item (HW6\underline{ }2)
         Let Y = outcome of a single octahedral (8 face) die.
	 $E(Y)=\displaystyle\sum_{x=1}^{8} { x \times P(X=x)} =
	   4.5$, And $Var(Y) = {1 \over N-1} \displaystyle\sum_{x=1}^{8} { 
		(x - E(Y))^2} = (approx.) 6$.

	 Then, if X = outcome of 6 dice rolled twice (total of
	   12 independent observations from the same distribution), we have
	 $E(X)=E(\displaystyle\sum_{k=1}^{12} { Y_k } ) =
		\displaystyle\sum_{k=1}^{12} { E(Y_k)} = 12 \times 4.5 = 54.$

	 $Var(X) = Var(\displaystyle\sum_{k=1}^{12} { Y_k }) = 
		\displaystyle\sum_{k=1}^{12} { Var(Y_k)} = 12 \times 6 = 72$.

	 $SD(Y) = 6 \sqrt{2}$. 


	 Let, now, $\overline{Y}$ be the average of 11 such experiments. Then
	 $\overline{Y} = {1 \over 11} \displaystyle\sum_{k=1}^{11} {X_k}$.
	 $E(\overline{Y}) = E(X) = 54$ and
	 $Var(\overline{Y}) = Var({1 \over 11}
		\displaystyle\sum_{k=1}^{11} { X_k }) = 
                {1 \over 11^2}\displaystyle\sum_{k=1}^{11} 
		{ Var(X_k)} = {1 \over 11^2} \times 11 \times 72$.
	 Lastly, $SD(\overline{Y}) = \sqrt{Var(\overline{Y})} = \sqrt{72/11}$.

         The distribution of sample average, $\bar{Y}$ is normal,
	 $\overline{Y} \sim N(54, 72/11)$.
    
  \item (HW6\underline{ }3)
         \begin{itemize}
         \item[(a)] Let $\bar{X_{1}}$, $S_{1}$, and $n_{2}$ be the mean,
                    SD and the sample size of the mnemonic group, respectively.
                    \begin{eqnarray*}
                    \bar{X_{1}}=\frac{\sum X_{1i}}{n_1}=14.1\\
                    S_{1}=\sqrt{\frac{\sum(X_{1i}-\bar{X_1})^2}{n_1-1}}=2.47
                    \end{eqnarray*}
                    Let $\bar{X_{2}}$, $S_{2}$, and $n_2$ be the mean, SD and the sample size
                    of the control group, respectively.
                    \begin{eqnarray*}
                    \bar{X_{2}}=\frac{\sum X_{2i}}{n_2}=9.22\\
                    S_{2}=\sqrt{\frac{\sum(X_{2i}-\bar{X_2})^2}{n_2-1}}=2.90
                    \end{eqnarray*} 
         \item[(b)] A 95\% C.I. for the difference in the mean
         number of words recalled between the normal and mnemonic
         instruction methods is
                    \begin{eqnarray*}
                    \bar{X_1}-\bar{X_2}\pm t
                    \sqrt{\frac{S_1^2}{n_1}+\frac{S_2^2}{n_2}}
                    \end{eqnarray*}
                    \begin{eqnarray*}
                    &=&14.1-9.22\pm
                    2.11\sqrt{\frac{2.47^2}{20}+\frac{2.9^2}{18}}\\
                    &=&[3.03,6.73].
                    \end{eqnarray*}
                    where $df=min\{20-1,18-1\}=17$,
                    $t_{17,0.25}=2.11$.

                    We have 95\% confident that the difference in the mean
         number of words recalled between the normal and mnemonic
         instruction methods is between 3.03 to 6.73.
     
         \item[(c)] To approximately halve the size of the CI we need to
quadruple the (largest) sample size. That is we'll need about 20x4 = 80
	  observations in every group.

         \item[(d)] The C.I. above may or may not contain the true difference
in means between the two groups. We have no knowledge about the true population
differences between the control and mnemonic groups. We only have a sample
\it{estimate} of this difference. Hence, with the given knowledge, we can not
affirmatively answer, or reject the fact that the true population difference
is in the interval. The only statement we can make is that \it{we are 95% sure
the true groups difference in memory performance is within the range of our
CI(95\%)}.

         \end{itemize}
       
\end{itemize}

\end{document}
